Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel: 4h2 o(g) + 3fe(s) ⟶ fe3 o4 (s) + 4h2 (g) (a) outline the steps necessary to answer the following question: what volume of h2 at a pressure of 745 torr and a temperature of 20 °c can be prepared from the reaction of 15.o g of h2o? (b) answer the question.

Answer : The volume of will be, 0.2690 L

Solution :

(a) Steps involved for this problem are :

First we have to calculate the moles of .

Now we have to calculate the volume of hydrogen gas by using the ideal gas equation.

(b) First we have to calculate the moles of .

The balanced chemical reaction is,

From the balanced chemical reaction, we conclude that the moles of hydrogen is equal to the moles of water.

Thus, the moles of hydrogen gas = 0.833 mole

Now we have to calculate the volume of hydrogen gas.

Using ideal gas equation,

where,

n = number of moles of gas  = 0.833 mole

P = pressure of the gas =

conversion used : (1 atm = 760 torr)

T = temperature of the gas =

R = gas constant = 0.0821 Latm/moleK

V = volume of gas = ?

Now put all the given values in the above equation, we get :

Therefore, the volume of will be, 0.2690 L

V = 20.4 L

Explanation:

Step 1: Write the balanced equation

4 H₂O(g) + 3 Fe(s) ⟶ Fe₃O₄(s) + 4 H₂(g)

Step 2: Find the moles of H₂

We can establish the following relations.

The moles of H₂ obtained from 15.0 g of H₂O is:

Step 3: Find the volume of H₂

We will use the ideal  gas equation.

P = 745 torr × (1 atm/760 torr) = 0.980 atm

V = ?

n = 0.833 mol

R = 0.08206 atm.L/mol.K

T = 20°C + 273 = 293 K

P × V = n × R × T

0.980 atm × V = 0.833 mol × (0.08206 atm.L/mol.K) × 293 K

V = 20.4 L

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